Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int N = 100, MOD = 1e9;
using namespace std;
int n, m, g[N + 9][N + 9], op[N + 9][N + 9], f;
queue<pair<bool, int> > q;
bool ok() {
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      if (op[i][j] != g[i][j]) return 0;
  return 1;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) scanf("%d", &g[i][j]);
  for (int i = 0; i < n && n <= m; ++i) {
    int minr = MOD;
    for (int k = 0; k < m; ++k) minr = min(minr, g[i][k]);
    for (int k = 0; k < m; ++k) op[i][k] = minr;
    for (int k = 0; k < minr; ++k) q.push({0, i + 1});
    f = 1;
  }
  for (int i = 0; i < m; ++i) {
    int minc = MOD;
    for (int k = 0; k < n; ++k) minc = min(minc, g[k][i] - op[k][i]);
    for (int k = 0; k < n; ++k) op[k][i] += minc;
    for (int k = 0; k < minc; ++k) q.push({1, i + 1});
  }
  for (int i = 0; i < n && !f; ++i) {
    int minr = MOD;
    for (int k = 0; k < m; ++k) minr = min(minr, g[i][k] - op[i][k]);
    for (int k = 0; k < m; ++k) op[i][k] += minr;
    for (int k = 0; k < minr; ++k) q.push({0, i + 1});
  }
  if (ok()) {
    printf("%d\n", (int)q.size());
    while (!q.empty()) {
      printf("%s %d\n", (q.front().first ? "col" : "row"), q.front().second);
      q.pop();
    }
  } else
    puts("-1");
  return 0;
}